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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>First we review some results in mechanics.1. Newton’s second law:This law gives <span class="process-math">\(F=ma\text{,}\)</span> where <span class="process-math">\(F\)</span> is the external force, <span class="process-math">\(m\)</span> is the mass and <span class="process-math">\(a\)</span> is the acceleration. The basic units are <span class="process-math">\(s\)</span> for time, <span class="process-math">\(m\)</span> for length and <span class="process-math">\(kg\)</span> for mass. The unit for the force is <span class="process-math">\(N\)</span> and for the acceleration is <span class="process-math">\(m/s^2\text{.}\)</span>2. Gravity of earthGravity is given by the formula</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textrm{Gravity}=-\frac{mg R^2}{(R+x)^2}=-\frac{mg}{\left(1+\frac{x}{R}\right)^2},
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(g=9.8 m /s^2\text{.}\)</span> For small <span class="process-math">\(x\text{,}\)</span> we have Taylor expansion of <span class="process-math">\((1+\frac{x}{R})^{-2}\)</span> so that</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\textrm{Gravity}\approx -mg (1-2 \frac{x}{R}+\cdots)\approx -mg.
\end{equation*}
</div>
<p class="continuation">3. Other external force For example, the air resistance can be regarded as being proportional to the velocity, say <span class="process-math">\(-kv\)</span> where <span class="process-math">\(k\)</span> is a known constant.<dfn class="terminology">Example 1</dfn> A body of constant mass <span class="process-math">\(m\)</span> is projected vertically upward from the surface of the earth with an initial velocity <span class="process-math">\(v_0\text{.}\)</span> The gravitational acceleration of the earth is assumed to be constant. During the motion, the body is subjected to an air resistance which is proportional to the magnitude of the velocity, say, <span class="process-math">\(k |v|\text{.}\)</span> Finda) The time at which the maximum height is reached;b) The maximum height attained by the body.<dfn class="terminology">Solution:</dfn> From Newton’s second law, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;F=-kv-mg=ma=m \frac{\textrm{d} v}{\textrm{d} t},\\
&amp; v(0)=v_0.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">We want to solve the above initial value problem. The differential equation is separable equation and it is rewritten as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
-\textrm{d} t=\frac{\textrm{d} v}{\frac{k}{m} v+g}.
\end{equation*}
</div>
<p class="continuation">Integrating on both sides gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
-t+C_1=\frac{m}{k} \ln |\frac{k}{m}v+g|~\rightarrow~|\frac{k}{m}v+g|=e^{\frac{k}{m} C_1} e^{-\frac{k}{m} t}~\rightarrow~\frac{k}{m} v+g=\pm e^{\frac{k}{m} C_1} e^{-\frac{k}{m} t}=C e^{-\frac{k}{m} t}.
\end{equation*}
</div>
<p class="continuation">So the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
v=\frac{m}{k}\left(C e^{-\frac{k}{m} t}-g \right).
\end{equation*}
</div>
<p class="continuation">Using the initial condition, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
v_0=\frac{m}{k} (C-g)~\rightarrow~C=\frac{k}{m} v_0+g.\tag{2.5.1}
\end{equation}
</div>
<p class="continuation">Therefore, the solution to the initial value problem is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
v=\frac{m}{k}\left[\left(\frac{k}{m} v_0+g\right) e^{-\frac{k}{m} t}-g \right]=\left(v_0+\frac{mg}{k} \right) e^{-\frac{k}{m} t}-\frac{mg}{k}.\tag{2.5.2}
\end{equation}
</div>
<span class="incontext"><a href="sec2_5.html#p-36" class="internal">in-context</a></span>
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